Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $z = \dfrac{9p + 72}{4p^2 + 28p} \div \dfrac{p - 1}{p^2 + 6p - 7} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{9p + 72}{4p^2 + 28p} \times \dfrac{p^2 + 6p - 7}{p - 1} $ First factor the quadratic. $z = \dfrac{9p + 72}{4p^2 + 28p} \times \dfrac{(p + 7)(p - 1)}{p - 1} $ Then factor out any other terms. $z = \dfrac{9(p + 8)}{4p(p + 7)} \times \dfrac{(p + 7)(p - 1)}{p - 1} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ 9(p + 8) \times (p + 7)(p - 1) } { 4p(p + 7) \times (p - 1) } $ $z = \dfrac{ 9(p + 8)(p + 7)(p - 1)}{ 4p(p + 7)(p - 1)} $ Notice that $(p - 1)$ and $(p + 7)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ 9(p + 8)\cancel{(p + 7)}(p - 1)}{ 4p\cancel{(p + 7)}(p - 1)} $ We are dividing by $p + 7$ , so $p + 7 \neq 0$ Therefore, $p \neq -7$ $z = \dfrac{ 9(p + 8)\cancel{(p + 7)}\cancel{(p - 1)}}{ 4p\cancel{(p + 7)}\cancel{(p - 1)}} $ We are dividing by $p - 1$ , so $p - 1 \neq 0$ Therefore, $p \neq 1$ $z = \dfrac{9(p + 8)}{4p} ; \space p \neq -7 ; \space p \neq 1 $